回答編集履歴
2
append answer
test
CHANGED
@@ -112,3 +112,50 @@
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112
112
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FFT: (4.000000 + 0.000000i) (0.000000 + 0.000000i) (0.000000 + 0.000000i) (0.000000 + 0.000000i) (4.000000 + 0.000000i) (0.000000 + 0.000000i) (0.000000 + 0.000000i) (0.000000 + 0.000000i)
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113
113
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IFFT: (1.000000 + 0.000000i) (0.000000 + 0.000000i) (1.000000 + 0.000000i) (0.000000 + 0.000000i) (1.000000 + 0.000000i) (0.000000 + 0.000000i) (1.000000 + 0.000000i) (0.000000 + 0.000000i)
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114
114
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```
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115
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+
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116
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+
```C:短く書く方
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117
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+
void _fft(double complex* x, int n, int inv) {
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118
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+
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119
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if (n == 1) {
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120
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+
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121
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return;
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122
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}
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123
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+
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124
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+
double complex* xe = malloc(n / 2 * sizeof(double complex));
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125
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+
double complex* xo = malloc(n / 2 * sizeof(double complex));
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126
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+
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127
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for (int i = 0; i < n / 2; i++) {
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128
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+
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129
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xe[i] = x[2 * i];
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130
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xo[i] = x[2 * i + 1];
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131
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}
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132
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+
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133
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+
_fft(xe, n / 2, inv);
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134
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+
_fft(xo, n / 2, inv);
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135
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+
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136
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for (int i = 0; i < n / 2; i++) {
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137
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+
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138
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double complex w = cexp((inv ? 1 : -1) * 2 * M_PI * I * i / n);
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139
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double complex t = w * xo[i];
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140
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+
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141
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x[i] = xe[i] + t;
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142
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x[i + n / 2] = xe[i] - t;
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143
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}
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144
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+
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145
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free(xe);
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146
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free(xo);
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147
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+
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148
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return;
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149
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}
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150
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+
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151
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void fft(double complex* x, int n) {
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152
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_fft(x, n, 0);
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153
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return ;
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154
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}
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155
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+
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156
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+
void ifft(double complex* x, int n) {
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157
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_fft(x, n, 1);
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158
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for (int i = 0; i < n; i++) x[i] /= n;
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159
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return ;
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160
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}
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161
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+
```
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1
fix answer
test
CHANGED
@@ -42,8 +42,7 @@
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42
42
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return;
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43
43
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}
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44
44
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45
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-
void ifft(double complex* x, int n) {
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45
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+
void _ifft(double complex* x, int n) {
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46
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-
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47
46
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if (n == 1) {
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48
47
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49
48
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return;
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@@ -58,8 +57,8 @@
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58
57
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xo[i] = x[2 * i + 1];
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59
58
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}
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60
59
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61
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-
fft(xe, n / 2);
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60
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+
_ifft(xe, n / 2);
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62
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-
fft(xo, n / 2);
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61
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+
_ifft(xo, n / 2);
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63
62
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64
63
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for (int i = 0; i < n / 2; i++) {
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65
64
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@@ -70,12 +69,16 @@
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70
69
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x[i + n / 2] = xe[i] - t;
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71
70
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}
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72
71
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73
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-
for (int i = 0; i < n; i++) x[i] /= n;
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74
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-
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75
72
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free(xe);
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76
73
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free(xo);
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77
74
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78
75
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return;
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76
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+
}
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77
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+
|
78
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+
void ifft(double complex* x, int n) {
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79
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_ifft(x, n);
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80
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+
for (int i = 0; i < n; i++) x[i] /= n;
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81
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+
return ;
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79
82
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}
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80
83
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81
84
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int main() {
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